## Saturday, August 23, 2014

### Problem Solutions (Part 2)

(1) Photo-electrons with minimum velocity imply that K min   applies here  so indicates the wavelength of incident light (l = 0.50 mm) is at the cutoff wavelength so: l = l c     Then the cutoff or threshold frequency is:

f c = c /l c =    (3 x 108 ms-1)/ (500 x  10-9 m) =  6 x 10 14 s-1

Then work function would be: f   = hf c

(6.62 x 10 -34 Js)   ( 6 x 10 14 s-1)

=   3.97  x 10 -19 J  =    (3.97  x 10 -19 J )/ 1.60  x 10 -19 J /eV) = 2.48 eV

(2) Stopping potential Vs = 0.54 V and l = 750 nm

f=  c/ l =   (3 x 108 ms-1)/ (750 x  10-9 m) =  4 x 10 14  s-1

The work function is obtained from: eVs =  hf  -  f

So, transposing:   f  =  hf -    eVs     =

(6.62 x 10 -34 Js)  (4 x 10 14 /s) -   (1.6 x 10 -19 C) 0.54 J/C]

f  =   2.64 x 10 -19 J  -   0.86 x  10 -19 J    = 1.78 x  10 -19 J

In eV:  f  =    1.78 x  10 -19 J / (1.6 x 10 -19 J/ eV) =  1.11 eV

(3) K max  =   0.57 eV   and photo-electrons dislodged from a metal surface by incident radiation with l = 500 nm.

The incident energy E = hf = h c/l  =

(6.62 x 10 -34 Js) (3 x 108 ms-1)/ (500 x  10-9 m)

h c/ l = 3.97  x 10-19 J

K max  =   0.57 eV (1.6 x 10 -19 J/ eV) = 0.91 x 10 -19 J

Therefore:

K max  =  0.91 x 10 -19 J

= [3.97  x 10-19 J  -   f ]

So, the work function is: f  =  [3.97  x 10-19 J    -  0.91 x 10 -19 J]  =  3.06 x 10 -19 J

To get in eV:   (3.06 x 10 -19 J)/ (1.6 x 10 -19 J/ eV) =  1.91 eV

The stopping potential in volts:

Vs =   hf/ e   -  f/e    where the slope h/e = 4.13 x 10-15 Js/C

The frequency f = c/l  =    (3 x 108 ms-1)/ (500 x  10-9 m) =

6 x 10 14 c/s

Then:

Vs = (4.13 x 10-15 J s/C )(6 x 10 14 s-1) -

(3.06 x 10 -19 J)/ (1.6 x 10 -19 C)

Vs   = 2.47 V – 1.91 V =  0.56 V